Derivative of e^(sin x): Answer, Proof, Mistakes

How to differentiate e^(sin x)

The outer function is $e^u$ and the inner is $u = \sin x$. The exponential is its own derivative, so differentiating the outer just reproduces $e^{\sin x}$; then multiply by the inner derivative $u' = \cos x$.

This is the general pattern $\frac{d}{dx}\,e^{g(x)} = g'(x)\,e^{g(x)}$: the exponential comes along unchanged and picks up the derivative of the exponent as a factor.

Reading the sign of the slope

Because $e^{\sin x} > 0$ for every $x$, the factor that controls whether the curve rises or falls is $\cos x$. The slope is zero exactly where $\cos x = 0$, at $x = \frac{\pi}{2} + n\pi$, which are the peaks and troughs of $\sin x$ and therefore of $e^{\sin x}$.

So $e^{\sin x}$ increases where $\cos x > 0$ and decreases where $\cos x < 0$, matching the increase and decrease of the exponent $\sin x$.

Common mistakes

• Answering $e^{\sin x}$ alone and dropping the chain rule factor $\cos x$. The inner function is $\sin x$, not $x$, so its derivative must appear.
• Writing $e^{\cos x}$, as if differentiating $\sin x$ changes the exponent inside the exponential. The exponent stays $\sin x$; the $\cos x$ multiplies out front.
• Using the product rule. There is no product here, just a composite, so the chain rule is the right tool.
• Answering $\cos x\,e^{\cos x}$ by differentiating the exponent in two places at once. Only the multiplier changes to $\cos x$; the exponent is untouched.